for what values of k is f continuous at x = 6? f(x) = { (x^2 - 2x - 24)/(x - 6), x ≠ 6; k, x = 6 }

Explanation:

Step1: Simplify the non - constant part of the function

For $x\neq6$, simplify $\frac{x^{2}-2x - 24}{x^{2}-5x - 6}$. Factor the numerator $x^{2}-2x - 24=(x - 6)(x + 4)$ and the denominator $x^{2}-5x - 6=(x - 6)(x+1)$. Then $\frac{x^{2}-2x - 24}{x^{2}-5x - 6}=\frac{(x - 6)(x + 4)}{(x - 6)(x + 1)}=\frac{x + 4}{x + 1}$ for $x\neq6$.

Step2: Use the definition of continuity

A function $y = f(x)$ is continuous at $x=a$ if $\lim_{x\rightarrow a}f(x)=f(a)$. We want $\lim_{x\rightarrow6}\frac{x^{2}-2x - 24}{x^{2}-5x - 6}=k$. Substitute $x = 6$ into $\frac{x + 4}{x + 1}$ (since the limit as $x\rightarrow6$ of $\frac{x^{2}-2x - 24}{x^{2}-5x - 6}$ is the same as the limit of the simplified form). So $\lim_{x\rightarrow6}\frac{x + 4}{x + 1}=\frac{6+4}{6 + 1}=\frac{10}{7}$.

Answer:

$\frac{10}{7}$